Quadratic Equations Test - 14

Hint:- Quadratic equation is always two roots. Quadratic equation makes a curve in two axis. Standard form of a quadratic equation.

$a{x}^2+bx+c=0$   [$a$ can't be zero]

Solution:-

Given one root is $3$.

Put $x=3$ in all parts

$x^2-5x+6=0$

$x=3$

${\left(3\right)}^2-5\left(3\right)+6=0$

$9-15+6=0$

$\therefore 0=0$

It means LHS=RHS

Hence, $x^2-5x+6=0$ is a required equation which has root $x=3$.

Note:- Always remember the power of $x$ in the equation and find the root.

Hence, the correct option is (A).

Hint:- Product is the multiplication of two numbers and sum is the addition of two numbers.

Solution:-

Given, the sum of the two numbers is $27$.

Product of these two numbers is $182$.

Let $x$ be the one number and the second one is $\left(27-x\right)$.

So, $x\left(27-x\right)=182$

$27x-x^2=182$

$x^2-27x+182=0$

It is a quadratic equation.

Now, 

$x^2-\left(14+13\right)x+182=0$

$x^2-14x-13x+182=0$

$x\left(x-14\right)-13\left(x-14\right)=0$

$\left(x-13\right)\left(x-14\right)$

If $x-13=0$

$\therefore x=13$

If $x-14=0$

$\therefore x=14$

Hence, the two numbers are $13$ and $14$.

Note:- Crosscheck,

$\left(13+14\right)=27$  (Sum)

$\left(13\times 14\right)=182$  (Product)

Hence, the correct option is (B).

Hint:- Remember the general form of a quadratic equation.

$x^2-\left(M+N\right)x-MN=0$

Here, $M$ and $N$ are the roots of the equation.

$\left[\left(a+b\right)\left(a-b\right)=\left(a^2-b^2\right)\right]$

Solution:-

Given the roots of a quadratic equation.

Let one root $M=7+\sqrt{3}$

And $2^{nd}$ one is $N=7-\sqrt{3}$

We know the general form of quadratic equation,

$x^2-\left(M+N\right)x+MN=0$

Put the value of $M$ and $N$ and solve them. We get a quadratic equation.

$x^2-\left(7+\sqrt{3}+7-\sqrt{3}\right)x+\left[\left(7+\sqrt{3}\right)\times \left(7-\sqrt{3}\right)\right]=0$

$x^2-14x+\left(49-3\right)=0$

If $\left(a+b\right)\left(a-b\right)$ i.e., it's equal to $\left(a^2-b^2\right)$.

So the quadratic equation is $x^2-14x-46=0$

Note:- Addition of roots is put with $b$ and multiplication with $c$.

$a{x}^2+bx+c=0$

Hence, the correct option is (A).

Hint:- Right-angle means $90°$

$\left[{\left(AB\right)}^2+{\left(BC\right)}^2={\left(AC\right)}^2\right]$  (Pythagoras theorem)

Solution:-

Let the shortest side is $x$.

Hypotenuse is $6$ more than twice of shortest side i.e., $2x+6$

The third side is $2$ meters less than the hypotenuse i.e., $2x+6-2$

$\therefore 2x+4$

It is a right-angle triangle so apply Pythagoras theorem,

${\left(AB\right)}^2+{\left(BC\right)}^2={\left(AC\right)}^2$

${\left(2x+4\right)}^2+{\left(x\right)}^2={\left(2x+6\right)}^2$

Note:- Apply Pythagoras theorem carefully.

Hence, the correct option is (A).

Hint:- Both are simple quadratic equation so find the roots and compare.

Solution:-

Given two quadratic equation,

$2x^2+x-6=0$...(1)

$x^2-3x-10=0$...(2)

Take equation (1) and find roots,

$2x^2+4x-3x-6=0$

$\Rightarrow 2x\left(x+2\right)-3\left(x+2\right)=0$

$\left(2x-3\right)\left(x+2\right)=0$

So roots are $\frac{3}{2}$ and $-2$.

Take equation (2),

$x^2-5x+2x-10=0$

$x\left(x-5\right)+2\left(x-5\right)=0$

$\left(x-5\right)\left(x+2\right)=0$

Roots are $5$ and $-2$.

$-2$ are common.

Note:- Solve quadratic equation carefully.

Hence, the correct option is (D).

Hint:- Two consecutive integers mean $a$ and $\left(a+1\right)$ and product is $a\left(a+1\right)=$?

Solution:-

Given the product of two consecutive integers is $240$.

Let integer is $X$ and $\left(X+1\right)$.

So the product is $X\left(X+1\right)=240$

Note:- Consecutive means that just next.

Hence, the correct option is (B).

Hint:- We know that the quadratic equation is two root equations. So to find the root solve the equation. Quadratic equation has $2$-degree equation because the power of $x$ is,

Solution:- 

Given, equation $a^2{x}^2-3abx+2b^2=0$

Break the equation and we get,

$a^2{x}^2-2abx-abx+2b^2=0$

$ax\left(ax-2b\right)-b\left(ax-2b\right)=0$

$\left(ax-b\right)\left(ax-2b\right)=0$

If $ax-b=0$

$\therefore x=\frac{b}{a}$

If $ax-2b=0$

$\therefore x=\frac{2b}{a}$

So the roots are $\frac{b}{a}$ and $\frac{2b}{a}$.

Note:- Put $x=\frac{2b}{a}$ in a given equation and solve we get LHS=RHS.

Hence, the correct option is (B).

Hint:- In a quadratic equation,

$a{x}^2+bx+c=0$

We know the value of $a, b, c$ so easily find the value of $x$ and $x$ has two roots means two value.

Solution:- Given, $x^2+px+q=0$...(1)

And $p=-7, q=12$

Put $p$ and $q$ value in equation (1) and solve,

$x^2-7x+12=0$

$x^2-\left(4+3\right)x+12=0$

$x^2-4x-3x+12=0$

Which gives $\left(x-3\right)\left(x-4\right)$

If $x-3=0$

$\therefore x=3$

And $x-4=0$

$\therefore x=4$

So the roots of a quadratic equation are $3$ and $4$.

Note:- Given equation is quadratic equation so it has two roots.

Hence, the correct option is (B).

Hint:- Quadratic equation is a $2$-degree equation.

$a{x}^2+bx+c=0$

Solution:-

Check by option (A),

$x^3-x^2={\left(x-1\right)}^3$

$x^3-x^2=x^3-1-3x^2+3x$

$-x^2+3x^2-3x+1=0$

$2x^2-3x+1=0$

Check by option (B),

$x^2+2x+1={\left(4-x\right)}^2+3$

$x^2+2x+1=16+x^2+8x+3$

$cx+18=0$

Not a quadratic equation because degree $=2$.

Check by option (C),

$\left(k+1\right)x^2+\frac{3}{2}x-5=0, k=-1$

$\left(-1+1\right)x^2+\frac{3}{2}x-5=0$

$\therefore \frac{3}{2}x-5=0$

Not a quadratic.

Check by option (D),

$-2x^2=\left(5-x\right)\left(2x-\frac{2}{5}\right)$

It is also not a quadratic equation.

Note:- This type of question solves by option step by step.

Hence, the correct option is (A).

Hint:- Use $\left[\left(a^2-b^2\right)=\left(a+b\right)\left(a-b\right)\right]$

Solution:-

Given, $x^2-4px+4p^2-q^2=0$

We know that ${\left(a-b\right)}^2=a^2+b^2-2ab$

Apply the following rule in the first term we get,

${\left(x-2p\right)}^2-q^2=0$

We know, 

$\left(a^2-b^2\right)=\left(a+b\right)\left(a-b\right)$

So,

$\left(x-20+q\right)\left(x-2p-q\right)=0$

If $x-2p+q=0$

$\therefore x=2p-q$

If $x-2p-q=0$

$\therefore x=2p+q$

So the roots are $2p-q$ and $2p+q$.

Note:- Always remember the quadratic equation has two roots.

Hence, the correct option is (D).

Hint:- If we have a number $a$ so it's reciprocal is $\frac{1}{a}$.

Solution:-

Let the number is $x$ and its reciprocal is $\frac{1}{x}$.

Given, $x+\frac{1}{x}=2$

$\frac{x^2+1}{x}=2$

$x^2-2x+1=0$

It is a quadratic equation so two roots is possible.

$x^2-x-x+1=0$

$x\left(x-1\right)-1\left(x-1\right)=0$

$\left(x-1\right)\left(x-1\right)=0$

If $x-1=0$

$\therefore x=1$

If $x-1=0$

$\therefore x=1$

$x+x=1+1=2$ (Sum)

And reciprocal is $\frac{1}{2}$

Note:- In these type of question always try to obtain a quadratic equation and solve.

Hence, the correct option is (C).

Hint:- If a given quadratic equation,

$a{x}^2+bx+c=0$

We know the value of $a, x$ and $c$ then put the all value in the equation and we get the value of $b$.

Solution:-

Given, $3x^2-px-2=0$ quadratic equation.

And $x=2$

Put $x=2$ in the equation,

$3x^2-px-2=0$

$3{\left(2\right)}^2-p^2-2=0$

$12-2p-2=0$

$10=2p$

$\therefore p=5$

Note:- If we cross-check the equation then put in equation $p=5$ in the given equation and we get two root's and one of them is $x=2$.

Hence, the correct option is (B).

Hint:- If $x$ is a variable or number then its reciprocal is $\frac{1}{x}$.

Solution:- Given the sum of two numbers is $17$ and sum of their reciprocal is $\frac{17}{62}$.

Let one number is $x$

Then second number is $\left(17-x\right)$

Because $17$ is the product of two numbers.

So the reciprocal is,

$\frac{1}{x}$ and $\frac{1}{\left(17-x\right)}$

Now given the sum of these reciprocals is $\frac{17}{62}$. So,

$\frac{1}{x}+\frac{1}{\left(17-x\right)}=\frac{17}{62}$...(1)

Note:- If we solve equation (1) further then we find its degree is $2$. Then, it is also a quadratic equation.

Hence, the correct option is (B).

Hint:- Quadratic equation is a $2$-degree equation,

$a{x}^2+bx+c=0$

Solution:-

To find the quadratic equation in given options.

By option (A),

${\left(\sqrt{2}x+\sqrt{3}\right)}^2+x^2=3x^2-5x$

$2x^2+3+2\sqrt{6}x+x^2=3x^2-5x$

That gives,

$2\sqrt{6}x+5x+3=0$

Not a quadratic equation because the maximum power of $x$ is $1$.

Note:- Degree of a quadratic equation is $2$.

Hence, the correct option is (A).

Hint:- We know that degree of a quadratic equation is $2$ and two roots.

General form of a quadratic equation.

$a{x}^2+bx+c=0$

Solution:-

Given equation, $5x^2+8x+4=2x^2+4x+6$

It is an equation where the degree is $2$.

Now,

$5x^2-2x^2+8x-4x+4-6=0$

$3x^2+4x-2=0$

This equation represents the quadratic equation because the degree is two $\left(2\right)$.

Note:- Always remember the degree of a quadratic equation is $2$.

Hence, the correct option is (C).